Determine the involution when given two pairs of point on a line

Question

I'm studying about involution on a projective line (line with point at infinity). An involution is a map from a projective line $l$ to itself that satisfied $f \circ f =$ is the identity map. According to the website http://users.math.uoc.gr/~pamfilos/eGallery/problems/InvolutionBasic.html , we can define an involution with two pair $(X_1, X_2), (Y_1, Y_2)$

The three pairs of points $(X_1,X_2), (Y_1,Y_2)$ and $(Z_1,Z_2)$ intercepted on (e) by corresponding pairs of opposite sides of the quadrilateral are in involution. This means that the involution $F$ on e defined by two pairs (out of the three), $(X_1,X_2), (Y_1,Y_2)$ say, interchanges also the points of the other pair $(Z_1,Z_2)$.

Based on their word:

To prove $F(Z_1) = Z_2$ it suffices to show the equality of cross ratios: $(Y_1,Y_2,X_1,Z_1)$ and $(Y_2,Y_1,X_2,Z_2)$.

, I suspect that they are saying that any map $F$ satisfies:

• $F$ maps $X_1$ to $X_2$ and reverse.
• $F$ maps $Y_1$ to $Y_2$ and reverse.
• $F$ maps an arbitrary point $P_1$ to point $P_2$ such that $(P_1,P_2, X_1,Y_1) = (P_2,P_1, X_2, Y_2)$.

is an involution. If so, I'm quite confused that for every point $P_1$, there can be two points $P_2$ satisfying $(P_1,P_2, X_1,Y_1) = (P_2,P_1, X_2, Y_2)$. If we choose $P_1$ to be the origin, let the coordinate of $P_2, X_1, Y_1, X_2, Y_2$ be $x, x_1, y_1, x_2, y_2$. Then the equation $\dfrac{x_1}{x_1 - x}: \dfrac{y_1}{y_1 - x} = \dfrac{x_2-x}{x_2}: \dfrac{y_2-x}{y_2}$. As $x \neq x_1, x_2, y_1, y_2$, the equivalent equation is $x_1x_2(y_1-x)(y_2-x) = y_1y_2(x_1-x)(x_2-x)$. The equation always has the root $x=0$ and it's easy to verify that the equation has another root other than $0$ (might be $\infty$). Thus $F$ can have a lot of fixed points!?

Or is it required that the $P_2$ must not be $P_1$ in the third condition of $F$? If so, $F$ always has no fixed point (which is contradictory also because there is involution has 2 fixed points).

Please explain me what is the involution defined with two pairs $(X_1, X_2), (Y_1, Y_2)$? Also, I would appreciate if you describe the involution without using homogeneous coordinates (maybe using the cross-ratio). This is because I want to explain the concept of involution defined with 4 points to my students in high school. Thank you.

Answer 1

I suspect that they are saying that any map $F$ satisfies:

• $F$ maps $X_1$ to $X_2$ and reverse.
• $F$ maps $Y_1$ to $Y_2$ and reverse.
• $F$ maps an arbitrary point $P_1$ to point $P_2$ such that $(P_1,P_2, X_1,Y_1) = (P_2,P_1, X_2, Y_2)$.

is an involution.

No, I don't understand them to say that.

One thing to get straight is the distinction between an arbitrary map, and a projective transformation. The latter will preserve all cross ratios. As I understand the text, it talks purely about maps which are projective transformations. Any such map on the real projective line is uniquely defined by knowing the images of three distinct points. For example, the transformation can be defined using

$$Y_1\mapsto Y_2,\quad Y_2\mapsto Y_1,\quad X_1\mapsto X_2\;.$$

This already implies $X_2\mapsto X_1$ and thus an involution. Why? Because a projective transformation that is not the identity has two fixed points. So $F$ has two fixed points, and $F^2$ also has the same two fixed points. But $F^2$ also has $Y_1$ and $Y_2$ as additional fixed points, which were not fixed points of $F$. Therefore $F^2$ must be the identity, and thus $F$ must be an involution.

In your statement, the first two bullet points, namely $F(X_1)=X_2, F(X_2)=X_1, F(Y_1)=Y_2$ and $F(Y_2)=Y_1$ are enough to make $F$ an involution, again under the assumption that $F$ is a projective transformation.

I don't know where you got the equation $(P_1,P_2, X_1,Y_1) = (P_2,P_1, X_2, Y_2)$ from. But if you pick $P_2=F(P_1)$ then this equation will hold for the involution defined above. So that's the second root of the equation you mentioned there. I would thus say that equation is a consequence of $F$ being an involution, not a condition.

What they are doing is use $(Y_1,Y_2,X_1,Z_1)$ and $(Y_2,Y_1,X_2,Z_2)$. So contrary to your equation using $P$, their equation uses three of the points that define the transformation in each of the cross ratios. This is important since as I said before, three points and their images uniquely define the transformation. So by including three points, they ensure that the equation is making a statement about this specific transformation and not any other.

As the page details:

$$(Y_1,Y_2,X_1,Z_1) \overset{(1)}= (F(Y_1),F(Y_2),F(X_1),F(Z_1)) \overset{(2)}= (Y_2,Y_1,X_2,F(Z_1)) \overset{(3)}= (Y_2,Y_1,X_2,Z_2)$$

Here (1) is due to the projective transformation preserving cross ratio. (2) is applying the definition of this specific transformation, where for some points you know the image. Then (3) is on showing that if $Z_1$ is indeed the one specific point of intersection, then $F(Z_1)$ is indeed $Z_2$, the matching other point of intersection. This equation (3) is then core of what the subsequent proof using perspectivities aims to show.

Please explain me what is the involution defined with two pairs $(X_1,X_2),(Y_1,Y_2)$?

This question leaves a lot of room for interpretation. What is any mathematical object anyway, if not the sum of its definitions. This involution is defined by $Y_1\mapsto Y_2, Y_2\mapsto Y_1, X_1\mapsto X_2$ but does that explain what it is?

But as you mention cross ratio, perhaps the following characterization can help you: $((X_1,X_2);(Y_1,Y_2);(Z_1,Z_2))$ is called a quadrilateral set if

$$(x_1-y_2)(y_1-z_2)(z_1-x_2)=(x_1-z_2)(y_1-x_2)(z_1-y_2)\;.$$

Ideally you'd use homogeneous coordinates, in which case each of the parenthesized differences in the equation would actually be a $2\times2$ determinant of homogeneous coordinates. But as you said you want to avoid homogeneous coordinates, I picked the formulation with the differences instead, close to your own handling of cross ratios.

This equation is true for the six points that gave rise to the definition of the involution. But since $Z_1,Z_2$ didn't actually get used in the definition of the involution, you can also turn it around as a characterization of the involution itself. The image of any point $P_1$ under the involution is the uniquely defined point $P_2$ for which $((X_1,X_2);(Y_1,Y_2);(P_1,P_2))$ form a quadrilateral set.

You can use that to turn it into a construction. Given $X_1,X_2,Y_1,Y_2,P_1$ on a line, construct an complete quadrilateral whose sides intersect the line in these five points so that matching letters correspond to opposite sides. Then the sixth point of intersection will be $P_2=F(P_1)$. The quadrilateral you get for $P_1\not\in\{Z_1,Z_2\}$ must be distinct from the one you started with, but the involution for it will still be the same.

My knowledge of quadrilateral sets comes mostly from lectures by Jürgen Richter-Gebert, so his book Perspectives on Projective Geometry is what I would recommend for further studies. Any mistakes in this answer here are probably mine.

Answer 2

Now I'm able to explain the involution defined with 4 points by an arbitrary function to explain to my students.

Claim: On the line $l$, consider four points $A, A', B, B'$ ($B \neq B'$, $A$ doesn't necessary different from $A'$). Consider the transformation $f: l \to l$ that maps $A$ to $A'$, $B$ to $B'$, $B'$ to $B$, and any point $C$ to $C'$ such that $(CA, BB') = (C'A', B'B)$. Then $f$ is an involution on $l$.

Proof. Consider two cases:

1. Case 1: $f(\infty) = \infty$. In this case, \begin{align*} & (\infty, A; B, B') = (\infty, A'; B', B) \\ & \Rightarrow \frac{\overline{B'A}}{\overline{BA}} = \frac{\overline{BA'}}{\overline{B'A'}} \Rightarrow \overline{B'A} \cdot \overline{B'A'} = \overline{BA} \cdot \overline{BA'}. \\ \end{align*} Thus, $B'$ and $B$ have the same power wrt the circle with diameter $AA'$, implying that $B$ is the reflection of $B'$ wrt the midpoint of $AA'$.

   For any $C$ with $C' = f(C)$, we have \begin{align*} & (C, A; B, B') = (C', A'; B', B) \\ & \Rightarrow \frac{\overline{BC}}{\overline{BA}} : \frac{\overline{B'C}}{\overline{B'A}} = \frac{\overline{B'C'}}{\overline{B'A'}} : \frac{\overline{BC'}}{\overline{BA'}}. \end{align*} Combining this with $BB'$ and $AA'$ having the same midpoint, we obtain $\overline{BC} \cdot \overline{BC'} = \overline{B'C} \cdot \overline{B'C'}$, which means that $B$ and $B'$ have the same power wrt the circle with diameter $CC'$. Hence, $C$ is the reflection of $C'$ through the midpoint of $BB'$. Thus, $f$ is the reflection about the midpoint of $AA'$ (which is an involution).

2. Case 2: $f(\infty) = O \neq \infty$. We have \begin{align*} & (\infty, A; B, B') = (O, A'; B', B) \\ & \Rightarrow \frac{\overline{B'A}}{\overline{BA}} = \frac{\overline{B'O}}{\overline{B'A'}} : \frac{\overline{BO}}{\overline{BA'}} \\ & \Rightarrow \frac{\overline{OB}}{\overline{OB'}} = \frac{\overline{BA}}{\overline{B'A}} \cdot \frac{\overline{BA'}}{\overline{B'A'}} \\ & \Rightarrow \frac{\overline{OB}}{\overline{OB'}} : \frac{\overline{A'B}}{\overline{A'B'}} = \frac{\overline{BA}}{\overline{B'A}} \\ & \Rightarrow (B, B'; O, A') = \frac{\overline{BA}}{\overline{B'A}} \\ & \Rightarrow (B, O; B', A') = 1 - \frac{\overline{BA}}{\overline{B'A}} = \frac{\overline{B'B}}{\overline{B'A}} \\ & \Rightarrow \frac{\overline{B'B}}{\overline{B'O}} : \frac{\overline{A'B}}{\overline{A'O}} = \frac{\overline{B'B}}{\overline{B'A}} \\ & \Rightarrow \frac{\overline{A'O}}{\overline{B'O}} = \frac{\overline{A'B}}{\overline{B'A}} = \frac{\overline{A'O} - \overline{A'B}}{\overline{B'O} - \overline{B'A}} = \frac{\overline{BO}}{\overline{AO}} \\ & \Rightarrow \overline{OA} \cdot \overline{OA'} = \overline{OB} \cdot \overline{OB'}. \end{align*} For any $C$ with $C' = f(C)$, we have $(C, A; B, B') = (C', A'; B', B)$. This implies that $$\frac{\overline{BA}}{\overline{B'A}} \cdot \frac{\overline{BA'}}{\overline{B'A'}} = \frac{\overline{BC}}{\overline{B'C}} \cdot \frac{\overline{BC'}}{\overline{B'C'}}.$$ Therefore, $$\frac{\overline{OB}}{\overline{OB'}} = \frac{\overline{BC}}{\overline{B'C}} \cdot \frac{\overline{BC'}}{\overline{B'C'}}.$$ Proceeding similarly as above, we get $\overline{OA} \cdot \overline{OA'} = \overline{OC} \cdot \overline{OC'}.$

   Thus, $f$ is an inversion with center $O$ and radius $\overline{OA} \cdot \overline{OA'}$ and is also an involution.